I am trying to prove that the sequence $ \left \lfloor{ \frac{ 10 + n }{ 2n } }\right \rfloor $converges to 0 by using the episilon method.
I tend to work backwards. So far, I've managed to do the following: \begin{align} |\left \lfloor{ \frac{ 10 + n }{ 2n } }\right \rfloor - 0 | &= |\left \lfloor{ \frac{ 10 + n }{ 2n } }\right \rfloor | \\ &\leq |\left \lfloor{ \frac{ 10 + n }{ 2n } }\right \rfloor | \\ &\leq | \frac{ 10 + n }{ 2n } | \\ &= \frac{ 10 + n }{ 2n } \\ &\leq \frac{ 10 + n }{ n } \\ &= \frac{ 10 }{ n } + 1 \\ \end{align}
Then, I took $ \epsilon > \frac{ 10 }{ n } + 1 $, which lead to having $ n > \frac{ 10 }{ \epsilon - 1 } $. Thus, it is sufficient to take $ N > \frac{ 10 }{ \epsilon - 1 } $ to guarantee that for all $ n \geq N $, $ |\left \lfloor{ \frac{ 10 + n }{ 2n } }\right \rfloor - 0 | < \epsilon $.
The problem I am having is that the previous must hold for all $ \epsilon > 0 $, including $ \epsilon = 1 $. It can be shown manually that for $ n \leq 11 $, $ |\left \lfloor{ \frac{ 10 + n }{ 2n } }\right \rfloor - 0 | < 1 $, so the way I defined $ N $ is not completely incorrect.
I am wondering if there is a way to define $ N $ so that I don't have to consider various cases.
Of course, if $n>10$, then $\left\lfloor\frac{10+n}{2n}\right\rfloor=0.$ So for any $\epsilon>0$, if $n>10$, then $$\left|\left\lfloor\frac{10+n}{2n}\right\rfloor-0\right|=0<\epsilon.$$ In other words $N_\epsilon=10$. This proves that the limit is $0$.
The reason your first method does not work is that your upper bounds are too loose. The floor function is critical for the limit being $0$, so you cannot just throw it away. Similarly, $2$ in the denominator should not be discarded.